LAPLACE TRANSFORMS

INTRODUCTION

There are techniques for finding the system response of a system described by a differential equation, based on the replacement of functions of a real variable (usually time or distance) by certain frequency-dependent representations, or by functions of a complex variable dependent upon frequency. The equations are converted from the time or space domain to the frequency domain through the use of mathematical transforms.

The Laplace Transformation

Differentialequations

Inputexcitation e(t)

Outputresponse r(t)

Time Domain Frequency Domain

Algebraicequations

Inputexcitation E(s)

Outputresponse R(s)

Laplace Transform

Inverse Laplace Transform

THE LAPLACE TRANSFORM

Let f(t) be a real function of a real variable t (time) defined for t>0. Then

is called the Laplace transform of f(t). The Laplace transform is a function of a complex variable s. Often s is separated into its real and imaginary parts: s=+j , where and are real variables.

After a solution of the transformed problem has been obtained in terms of s, it is necessary to "invert" this transform to obtain the solution in terms of the time variable, t. This transformation from the s-domain into the t-domain is called the inverse Laplace transform.

THE INVERSE LAPLACE TRANSFORM

Let F(s) be the Laplace transform of a function f(t), t>0. The contour integral

where c>0 (0 as above) is called the inverse Laplace transform of F(s).

It is seldom necessary to perform the integration in the Laplace transform or the contour integration in the inverse Laplace transform. Most often, Laplace transforms and inverse Laplace transforms are found using tables of Laplace transform pairs.

Functional Laplace Transform Pairs

Time Domainf(t), t>0

Frequency DomainF(s)

1. 12. K K/s3. Kt K/s2

4. Ke-at K/(s+a)5. Kte-at K/(s+a)2

6. Ksint K/(s2+2)

7. Kcost Ks/(s2+2)8. Ke-atsint K/((s+a)2+2))

9. Ke-atcost K(s+a)/((s+a)2+2))

Operational Laplace Transform Pairs

Time Domain f(t), t>0

Frequency Domain F(s)

10. t s 11. f(t) F(s) 12. L-1{F(s)}=f(t) L{f(t)}=F(s) 13. Af1(t) + Bf2(t) AF1(s)+BF2(s)

14. t

df0

)(

F s

s

( )

15. df t

dt

( )

sF s f( ) ( ) 0

Inverse Laplace Transform

The inverse Laplace transform is usually more difficult than a simple table conversion.

X ss s

s s s( )

( )( )

( )( )

8 3 8

2 4

Partial Fraction Expansion

If we can break the right-hand side of the equation into a sum of terms and each term is in a table of Laplace transforms, we can get the inverse transform of the equation (partial fraction expansion).

X ss s

s s s

K

s

K

s

K

s( )

( )( )

( )( )

8 3 8

2 4 2 41 2 3

Repeated Roots

In general, there will be a term on the right-hand side for each root of the polynomial in the denominator of the left-hand side. Multiple roots for factors such as (s+2)n will have a term for each power of the factor from 1 to n.

Y ss

s

K

s

K

s( )

( )

( ) ( )

8 1

2 2 221 2

2

Complex Roots

Complex roots are common, and they always occur in conjugate pairs. The two constants in the numerator of the complex conjugate terms are also complex conjugates.

Z ss s

K

s j

K

s j( )

.

( ) ( )

*

52

2 5 1 2 1 22

where K* is the complex conjugate of K.

Solution of Partial Fraction Expansion The solution of each distinct (non-multiple)

root, real or complex uses a two step process. The first step in evaluating the constant is to

multiply both sides of the equation by the factor in the denominator of the constant you wish to find.

The second step is to replace s on both sides of the equation by the root of the factor by which you multiplied in step 1

X ss s

s s s

K

s

K

s

K

s( )

( )( )

( )( )

8 3 8

2 4 2 41 2 3

Ks s

s s s

1

0

8 3 8

2 4

8 0 3 0 8

0 2 0 424

( )( )

( )( )

( )( )

( )( )

Ks s

s s s

2

2

8 3 8

4

8 2 3 2 8

2 2 412

( )( )

( )

( )( )

( )

Ks s

s s s

3

4

8 3 8

2

8 4 3 4 8

4 4 44

( )( )

( )

( )( )

( )

The partial fraction expansion is:

X ss s s

( )

24 12

2

4

4

The inverse Laplace transform is found from the functional table pairs to be:

x t e et t( ) 24 12 42 4

Repeated Roots

Any unrepeated roots are found as before. The constants of the repeated roots (s-a)m

are found by first breaking the quotient into a partial fraction expansion with descending powers from m to 0:

B

s a

B

s a

B

s am

m( ) ( ) ( )

2

21

The constants are found using one of the following:

BP a

Q s s am m

s a

( )

( ) / ( )

1)/()(

)(

)!(

1

1 as

mim

im

i assQ

sP

ds

d

imB

Y ss

s

K

s

K

s( )

( )

( ) ( )

8 1

2 2 221 2

2

Ks s

ss

ss2

2

22

2

8 1 2

28 1 8

( )( )

( )( )

The partial fraction expansion yields:

Y ss s

( )( )

8

2

8

2 2

8)2/()2(

)1(8

)!12(

1

222

s

i ss

s

ds

dB

The inverse Laplace transform derived from the functionaltable pairs yields:

y t e tet t( ) 8 82 2

A Second Method for Repeated Roots

Y ss

s

K

s

K

s( )

( )

( ) ( )

8 1

2 2 221 2

2

21 )2()1(8 KsKs

211 288 KKsKs

Equating like terms:

211 288 KKandK

211 288 KKandK

2828 K

28168 K

Thus

22

8

2

8)(

sssY

tt teety 22 88)(

Another Method for Repeated Roots

221

2 22)2(

)1(8)(

s

K

s

K

s

ssY

Ks s

ss

ss2

2

22

2

8 1 2

28 1 8

( )( )

( )( )

As before, we can solve for K2 in the usual manner.

2212

22

2

8)2(

2)2(

)2(

)1(8)2(

s

ss

Ks

s

ss

ds

Ksd

ds

sd 82)1(8 1

18 K

22 2

8

2

8

)2(

)1(8)(

sss

ssY

tt teety 22 88)(

Unrepeated Complex Roots

Unrepeated complex roots are solved similar to the process for unrepeated real roots. That is you multiply by one of the denominator terms in the partial fraction and solve for the appropriate constant.

Once you have found one of the constants, the other constant is simply the complex conjugate.

Complex Unrepeated Roots

Z ss s

K

s j

K

s j( )

.

( ) ( )

*

52

2 5 1 2 1 22

Ks j

s j s jj

s j

5 2 1 2

1 2 1 213

1 2

. ( )

( )( ).

K j* . 13

)21(

3.1

)21(

3.1

52

2.5)(

2 js

j

js

j

sssZ

)21()21()(

3.13.1

js

e

js

esZ

jj

Case 1: Functions with repeated linear roots Consider the following example:

F(s) should be decomposed for Partial Fraction

Expansion as follows:

22) + 1)(s + (s

6s = F(s)

22 2 + s

C +

2 + s

B +

1 + s

A =

2) + 1)(s + (s

6s = F(s)

Using the residue method:

6

6

1 s

6

1 s

6s

ds

d F(s)2) (s

ds

d B

12 12-

1 s

6s F(s)2) (s C

6- 1

6-

2 s

6s 1)F(s) (s A

2- s2- s2-

2

2- s2-

2

21- s

1-

22 1

1

s

s

s

so

and f(t) = [-6e-t + (6 + 12t)e-2t ]u(t)

22 2 + s

12 +

2 + s

6 +

1 + s

6- =

2) + 1)(s + (s

6s = F(s)

Case 2: Functions with complex roots If a function F(s) has a complex pole (i.e., a

complex root in the denominator), it can be handled in two ways: 1) By keeping the complex roots in the form of a

quadratic 2) By finding the complex roots and using

complex numbers to evaluate the coefficients

Example: Both methods will be illustrated using the following example. Note that the quadratic terms has complex roots.

F(s) = 5s - 6s + 21

(s + 1)(s + 2s + 17)

2

2

Method 1: Quadratic factors in F(s) F(s) should be decomposed for Partial

Fraction Expansion as follows:

F(s) = 5s - 6s + 21

(s + 1)(s + 2s + 17) =

A

s + 1 +

Bs + C

s + 2s + 17

2

2 2

A) Find A, B, and C by hand (for the quadratic factor method): Combining the terms on the right with a

common denominator and then equating numerators yields:

A(s + 2s + 17) + (Bs + C)(s + 1) = 5s - 6s + 21 2 2

Equating s terms: A + B = 5

Equating s terms: 2A + B + C = - 6

Equating constants: 17A + C = 21

yields

A = 2

B = 3

C = -13

2

so

now manipulating the quadratic term into the form for decaying cosine and sine terms:

F(s) = 5s - 6s + 21

(s + 1)(s + 2s + 17) =

2

s + 1 +

3s - 13

s + 2s + 17

2

2 2

F(s) =

2

s + 1 +

3(s + 1)

s + 1 + 4 +

-4(4)

s + 1 + 42 2 2 2

so

The two sinusoidal terms may be combined if desired using the following identity:

f(t) = e + 3cos(4t) - 4sin(4t) u(t)-t 2

Acos(wt) + Bsin(wt) = A + B cos wt - tanB

A2 2 -1

so

f(t) = e + 5cos(4t + 53.13 ) u(t)-t 2

Method 2: Complex roots in F(s) Note that the roots of are

so

17) + 2s + (s 2

j4 1- = jw = s , s 21

F(s) = 5s - 6s + 21

(s + 1)(s + 2s + 17) =

5s - 6s + 21

(s + 1)(s + 1 - j4)(s + 1 + j4)

2

2

2

A) Find A, B, and C by hand (for the complex root method): F(s) should be decomposed for Partial

Fraction Expansion as follows:

F(s) = A

s + 1 +

s + 1 - j4 +

s + 1 + j4 where is a complex number

and is the conjugate of .

*B B

B

B B*

The inverse transform of the two terms with complex roots will yield a single time-domain term of the form

Using the Residue Theorem:

2 = 2B/ = 2Be cos(wt + ) tB

A = (s + 1)F(s) = 5s - 6s + 21

(s + 2s + 17) =

32

16 = 2

s = -1

2

2s = -1

B

B

B

= (s + 1 - j4)F(s) = 5s - 6s + 21

(s + 1)(s + 1 + j4)

= 5(-1,4) - 6(-1,4) + 21

(-1 + j4 + 1)(-1 + j4 + 1 + j4) = 2.5 /53.13

It is not necessary to also find but doing so here illustrates the conjugate relationship.

= (s + 1 + j4)F(s) = 5s - 6s + 21

(s + 1)(s + 1 - j4)

= 5(-1,-4) - 6(-1,-4) + 21

(-1 - j4 + 1)(-1 - j4 + 1 - j4)

s = -1 + j4

2

s = -1 + j4

2

*

s = -1 - j4

2

s = -1 - j4

2

*,

= 2.5 / - 53.13

So,

This can be broken up into separate sine and cosine terms using

f(t) = 2e u(t) + 2B/ = 2e u(t) + 5/53.13 -t -t

f(t) = e + 5e cos(4t + 53.13 ) u(t)-t -t2

cos(wt + ) = cos( )cos(wt) - sin( )sin(wt)

so

f(t) = e + 5e cos(53.13 )cos(4t) - sin(53.13 )sin(4t) u(t) -t -t2

f(t) = e + e 4cos(4t) - 3sin(4t) u(t)-t -t2